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bekas [8.4K]
4 years ago
13

Pleaseee help please!!! Picture is above :(

Mathematics
1 answer:
ololo11 [35]4 years ago
8 0

Answer:

The correct option is D) 11.4

Step-by-step explanation:

Given points are A(5,7) and B(-2,-2)

The distance between two points is given by

L=\sqrt{(X2-X1)^{2}+{(Y2-Y1)^{2}}

To find the distance between A(5,7) and B(-2,-2)

L=\sqrt{(X2-X1)^{2}+{(Y2-Y1)^{2}}

L=\sqrt{((-2)-5)^{2}+{((-2)-7)^{2}}

L=\sqrt{(-7)^{2}+{(-9)^{2}}

L=\sqrt{49+81}

L=11.4

Thus,

The correct option is D) 11.4

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It is known that the population variance equals 484. With a .95 probability, the sample size that needs to be taken if the desir
Ksju [112]

Answer:

We need a sample size of at least 75.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.025 = 0.975, so z = 1.96

Now, we find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

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The standard deviation is the square root of the variance. So:

\sigma = \sqrt{484} = 22

With a .95 probability, the sample size that needs to be taken if the desired margin of error is 5 or less is

We need a sample size of at least n, in which n is found when M = 5. So

M = z*\frac{\sigma}{\sqrt{n}}

5 = 1.96*\frac{22}{\sqrt{n}}

5\sqrt{n} = 43.12

\sqrt{n} = \frac{43.12}{5}

\sqrt{n} = 8.624

(\sqrt{n})^{2} = (8.624)^{2}

n = 74.4

We need a sample size of at least 75.

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